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一类非线性分数阶微分方程耦合系统正解的存在性

薛益民, 彭钟琪

薛益民, 彭钟琪. 一类非线性分数阶微分方程耦合系统正解的存在性[J]. 华南师范大学学报(自然科学版), 2020, 52(2): 102-106. DOI: 10.6054/j.jscnun.2020032
引用本文: 薛益民, 彭钟琪. 一类非线性分数阶微分方程耦合系统正解的存在性[J]. 华南师范大学学报(自然科学版), 2020, 52(2): 102-106. DOI: 10.6054/j.jscnun.2020032
XUE Yimin, PENG Zhongqi. On the Existence of Positive Solutions to the Coupled System of a Class of Nonlinear Fractional Differential Equations[J]. Journal of South China Normal University (Natural Science Edition), 2020, 52(2): 102-106. DOI: 10.6054/j.jscnun.2020032
Citation: XUE Yimin, PENG Zhongqi. On the Existence of Positive Solutions to the Coupled System of a Class of Nonlinear Fractional Differential Equations[J]. Journal of South China Normal University (Natural Science Edition), 2020, 52(2): 102-106. DOI: 10.6054/j.jscnun.2020032

一类非线性分数阶微分方程耦合系统正解的存在性

基金项目: 

国家自然科学基金项目 11526177

江苏省自然科学基金项目 BK20151160

徐州工程学院培育项目 XKY2017113

详细信息
    通讯作者:

    薛益民,副教授,Email:xueym@xzit.edu.cn

  • 中图分类号: O211.6

On the Existence of Positive Solutions to the Coupled System of a Class of Nonlinear Fractional Differential Equations

  • 摘要: 利用Guo-Krasnoselskii不动点定理、Schauder不动点定理和格林函数的性质,研究一类非线性Riemann-Liouville型分数阶微分方程耦合系统正解的存在性,得到了该耦合系统正解的存在性定理,并举例说明了定理的有效性.
    Abstract: The Guo-Krasnoselskii's fixed point theorem, the Schauder fixed point theorem and the properties of the associated Green's function are used to study the existence of positive solutions to the coupled system of a class of nonlinear Riemann-Liouville fractional differential equations. Two theorems about the existence of positive solutions are obtained, and two examples are given to illustrate the advantages of the theorems.
  • 随着非线性问题研究的深入,学者们建立了比整数阶微分方程模型更为精细的分数阶微分方程模型,以更好地解决复杂的实际问题,而其中的很多问题可以化为非线性分数阶微分方程边值问题.关于非线性分数阶微分方程边值问题正解的研究也已有许多重要成果[1-10].

    BAI和LU[10]利用锥上的不动点定理,得到了非线性分数阶边值问题

    {Dα0+u(t)+f(t,u(t))=0(0<t<1),u(0)=u(1)=0

    正解的存在性和多重性结果,其中,D0+α (1 < α≤2)表示Riemann-Liouville分数阶导数,fC([0, 1]×[0, ∞), [0, ∞)).

    受文献[10]的启发,本文利用Guo-Krasnoselskii不动点定理、Schauder不动点定理和格林函数的性质,得到如下的非线性Riemann-Liouville型分数阶微分方程耦合系统边值问题

    {Dαu(t)+f(t,v(t))=0(0<t<1),Dβv(t)+g(t,u(t))=0(0<t<1),u(0)=Dγu(0)=Dγu(1)=0,v(0)=Dδv(0)=Dδv(1)=0 (1)

    正解的存在性的充分条件,其中,2 < α, β≤3, 1 < γ≤2, 1 < δ≤2, 1+γα, 1+δβ, f, gC([0, 1]×[0, ∞), [0, ∞)), Dλ表示λ阶Riemann-Liouville分数阶导数,λ∈{α, β, γ, δ},并举例说明了定理的有效性.

    定义1[11]  函数f:R+Rα>0阶Riemann-Liouville积分为

    Iαf(t)=1Γ(α)t0(ts)α1f(s)ds,

    其中,等式右边在R+上逐点定义.

    定义2[11]  函数f:R+Rα>0阶Riemann-Liouville导数为

    Dαf(t)=1Γ(nα)(ddt)nt0f(s)(ts)αn+1ds,

    其中,n=[α]+1, [α]表示实数α的整数部分,等式右边在R+上逐点定义.

    引理1[11]  若α, β>0, f(x)∈L(0, 1), 则:

    (ⅰ) DβIαf(t)=Iαβf(t) (α>β);

    (ⅱ) DαIαf(t)=f(t);

    (ⅲ) IαDαf(t)=f(t)+ni=1citαi (ciR, i=1, 2, …, n, n-1 < αn, Dαf(t)∈L(0, 1));

    (ⅳ) Dαtγ=Γ(γ+1)Γ(γ+1α)tγα(γ>1,γ>α1,t>0).

    下面给出本文定理证明时所需要的引理.

    引理2[12]  (Guo-Krasnoselskii不动点定理)假设P在Banach空间E中是一个锥, Ω1Ω2E的有界开子集,且0∈Ω1, Ω1Ω2.如果A:P∩(Ω2\Ω1)→P是一个全连续算子,且下列条件之一成立:

    (ⅰ) ‖Ax‖≤ ‖x‖(xP∂Ω1)且‖Ax‖≥ ‖x‖(xPΩ2);

    (ⅱ)‖Ax‖≥ ‖x‖(xP∂Ω1)且‖Ax‖≤‖x‖(xPΩ2),

    AP∩(Ω2\Ω1)上有一个不动点.

    引理3[13]  (Schauder不动点定理)假设U是Banach空间X的非空有界闭凸子集,TU到其自身的全连续映射,则至少存在一个xU,使得Tx=x.

    引理4[14]  ∀y(t)∈C[0, 1], 2 < α≤3, 1 < γ≤2, 1+γα,分数阶微分方程边值问题

    {Dαu(t)+y(t)=0,0<t<1u(0)=Dγu(0)=Dγu(1)=0

    有唯一解u(t)=10Gα(t,s)y(s)ds, 其中

    Gα(t,s)={tα1(1s)αγ1(ts)α1Γ(α)(0st1),tα1(1s)αγ1Γ(α)(0ts1).

    类似可得

    Gβ(t,s)={tβ1(1s)βδ1(ts)β1Γ(β)(0st1)tβ1(1s)βδ1Γ(β)(0ts1)

    引理5[15]  假设G(t, s)=(Gα(t, s), Gβ(t, s)),则G(t, s)满足:

    (ⅰ) ∀t, s∈[0, 1],有G(t, s)∈C([0, 1]×[0, 1]);

    (ⅱ) ∀t, s∈[0, 1],有G(t, s)≥0,且∀t, s∈(0, 1),有G(t, s)>0;

    (ⅲ) ∀s∈[0, 1],有maxt[0,1]G(t,s)=G(1,s)

    (ⅳ) ∀s∈[0, 1],有

    mint[1/2,1]G(t,s)μmaxt[0,1]G(t,s)=μG(1,s),

    其中μ=min{μα=(1/2)α-1, μβ=(1/2)β-1}.

    X={u(t) |u(t)∈C([0, 1], [0, ∞))}. ∀uX,定义范数u=maxt[0,1]|u(t)|,则(X, ‖·‖)是Banach空间.令Y={v(t)|v(t)∈C([0, 1], [0, ∞))}. ∀vY,定义范数v=maxt[0,1]|v(t)|. ∀(u, v)∈X×Y,定义范数‖(u, v)‖ = ‖u‖+‖v‖,则(X×Y, ‖(u, v)‖)也是Banach空间.定义锥UX×Y

    U={(u(t),v(t))X×Y:u(t)0,v(t)0,t[0,1]}.

    ∀(u, v)∈X×Y,定义算子T:X×YX×Y

    T(u,v)(t)=(Tαv(t),Tβu(t))=(10Gα(t,s)f(s,v(s))ds,10Gβ(t,s)g(s,u(s))ds). (2)

    由引理4知T的不动点即为耦合系统(1)的解.

    引理6[14]  假设f, gC([0, 1]×[0, ∞), [0, ∞)),则算子T:UU为全连续的.

    为叙述简洁,记

    L=min{Lα=(10Gα(1,s)ds)1,Lβ=(10Gβ(1,s)ds)1},
    Δ=max{Δα=(11/2(1/2)α1Gα(1,s)ds)1,
    Δβ=(11/2(1/2)β1Gβ(1,s)ds)1},

    其中,μαμβ由引理5的(ⅳ)给出.

    定理1  假设f, gC([0, 1]×[0, ∞), [0, ∞)),若存在常数Ri>ri>0 (i=1, 2),使得下列不等式成立:

    (H1)f(t,v)M1R1((t,v)[0,1]×[0,R1]);
    (H2)f(t,v)M1r1((t,v)[0,1]×[0,r1]);
    (H3)g(t,u)M2R2((t,u)[0,1]×[0,R2]);
    (H4)g(t,u)N2r2((t,u)[0,1]×[0,r2]);
    (H5)0M1,M2L,N1,N2Δ,

    则耦合系统(1)至少有一个正解.

    证明  由引理6可知算子T:UU是全连续的.令

    ΩR={(u(t),v(t))|(u(t),v(t))X×Y,(u(t),v(t))<R=R1+R2,t[0,1]}.

    ∀(u, v)∈U∂ΩR,有‖(u, v)‖ =R. ∀t∈[0, 1],由(H1)、(H5)和引理5的(ⅱ)、(ⅲ),有

    Tαv(t)=10Gα(t,s)f(s,v(s))dsM1R110Gα(1,s)dsLαR110Gα(1,s)ds=R1,

    Tαv(t)R1. (3)

    由(H3)、(H5)和引理5的(ⅱ)、(ⅲ),有

    Tβu(t)=10Gβ(t,s)g(s,u(s))dsM2R210Gβ(1,s)dsLβR210Gβ(1,s)ds=R2,

    Tβu(t)R2. (4)

    由式(3)、(4),可得

    T(u,v)=Tαv(t)+Tβu(t)R1+R2=R=(u,v),

    T(u,v)(u,v)((u,v)UΩR).

    Ωr={(u(t), v(t))|(u(t), v(t))∈X×Y, ‖(u(t), v(t))‖ < r=r1+r2, t∈[0, 1]}. ∀(u, v)∈UΩr, 有‖(u, v)‖ =r. ∀t∈[0, 1],由(H2)、(H5)和引理5的(ⅱ)、(ⅳ),有

    Tαv(t)=10Gα(t,s)f(s,v(s))dsN1r110Gα(t,s)ds=N1r1(1/20Gα(t,s)ds+11/2Gα(t,s)ds)N1r111/2Gα(t,s)dsN1r111/2mint[1/2,1]Gα(t,s)dsΔαr111/2(1/2)α1Gα(1,s)ds=r1,

    Tαv(t)r1. (5)

    由(H4)、(H5)和引理5的(ⅱ)、(ⅳ),有

    Tβu(t)=10Gβ(t,s)g(s,u(s))dsN2r210Gβ(t,s)ds=N2r2(1/20Gβ(t,s)ds+11/2Gβ(t,s)ds)N2r211/2Gβ(t,s)dsN2r211/2mint[1/2,1]Gβ(t,s)dsΔβr211/2(1/2)β1Gβ(1,s)ds=r2,

    Tβu(t)r2. (6)

    由式(5)、(6),可得

    T(u,v)=Tαv(t)+Tβu(t)r1+r2=r=(u,v),

    T(u,v)(u,v)((u,v)UΩr).

    由引理2,算子T至少有一个不动点(u, v)∈U∩(ΩR\Ωr),即耦合系统(1)至少有一个正解.证毕.

    定理2  假设f, gC([0, 1]×[0, ∞), [0, ∞)),若存在常数ai, bi>0 (i=1, 2),使得下列不等式成立:

    (I1) f(t, v)≤a1+b1vρ1, ∀(t, v)∈[0, 1]×[0, ∞), ρ1∈(0, 1);

    (I2) g(t, u)≤a2+b2uρ2, ∀(t, u)∈[0, 1]×[0, ∞), ρ2∈(0, 1),

    则耦合系统(1)至少有一个正解.

    证明  取定常数r*,使得

    rmax{4a1Λ1,(4b1Λ1)11ρ1,4a2Λ2,(4b2Λ2)11ρ2}, (7)

    其中,Λ1=10Gα(1,s)ds,Λ2=10Gβ(1,s)ds.定义

    Ωr={(u,v)(t)|(u,v)(t)X×Y,(u,v)(t)r,t[0,1]}, (8)

    Ωr*是Banach空间X×Y的非空有界闭凸子集,下面证明T:Ωr*Ωr*. ∀(u, v)∈Ωr*,由式(2)、(7)、(8)和(I1),有

    Tαv(t)=10Gα(t,s)f(s,v(s))ds10Gα(1,s)(a1+b1vρ1)ds(a1+b1rρ1)Λ1r/4+r/4=r/2,

    Tαv(t)r/2. (9)

    由式(2)、(7)、(8)和(I2),有

    Tβu(t)=10Gβ(t,s)g(s,u(s))ds10Gβ(1,s)(a2+b2uρ2)ds(a2+b2rρ2)Λ2r/4+r/4=r/2,

    Tβu(t)r/2. (10)

    由式(9)、(10),可得

    T(u,v)=Tαv(t)+Tβu(t)r((u,v)Ωr),

    TΩr*Ωr*.

    由引理6,知算子TΩr*Ωr*是全连续的.由引理3,耦合系统(1)至少有一个正解.证毕.

    本节给出2个例子以验证定理的有效性.

    例1  考虑如下耦合系统边值问题

    {D5/2u(t)+t23+t2+tln(1+v(t))=0(0<t<1)D7/3v(t)+t2sint4+tu(t)(1+et)(1+u(t))=0(0<t<1)u(0)=D3/2u(0)=D3/2u(1)=0v(0)=D5/4v(0)=D5/4v(1)=0 (11)

    其中,2 < α=5/2, β=7/3≤3, 1 < γ=3/2, δ=5/4≤2,满足1+γα, 1+δβ,而且

    f(t,v(t))=v2+sin2t10+1,g(t,u(t))=v22+t+t3+1,

    易知f, gC([0, 1]×[0, ∞), [0, ∞)).经计算,可得

    10Gα(1,s)ds=1Γ(α)10((1s)αγ1(1s)α1)ds=1Γ(2.5)(10(1s)0ds10(1s)3/2ds)15.538918,
    10Gβ(1,s)ds=1Γ(β)10((1s)βλ1(1s)β1)ds=1Γ(7/3)(10(1s)1/12ds10(1s)4/3ds)15.691837.

    因此

    M=min{Mα=(10Gα(1,s)ds)1,Mβ=(10Gβ(1,s)ds)1}5.538918.

    由于

    11/2(1/2)α1Gα(1,s)ds=(1/2)α1Γ(α)11/2((1s)αγ1(1s)α1)ds=(1/2)3/2Γ(2.5)11/2((1s)0(1s)3/2)ds121.896336,
    11/2(1/2)β1Gβ(1,s)ds=(1/2)β1Γ(β)11/2((1s)βλ1(1s)β1)ds=(1/2)4/3Γ(2.5)11/2((1s)1/12(1s)4/3)ds123.886016.

    因此

    N=max{Nα=(11/2(1/2)α1Gα(1,s)ds)1,Nβ=(11/2(1/2)β1Gβ(1,s)ds)1}23.886016.

    选取R1=4, r1=1/12,R2=3, r2=1/9,则有

    f(t,v(t))=v+sin2t1+et+272R1=M1R1((t,v)[0,1]×[0,R1]),
    f(t,v(t))=v+sin2t1+et+22=24r1=N1r1((t,v)[0,1]×[0,r1]),
    g(t,u(t))=v22+t+t3+3114R2=M2R2((t,u)[0,1]×[0,R2]),
    g(t,u(t))=v22+t+t3+3325r2=N2r2((t,u)[0,1]×[0,r2]),

    且满足0 < M1, M2MN1, N2N.定理1的条件均被满足,所以耦合系统(11)至少存在1个正解.

    例2  考虑如下耦合系统边值问题

    {cD5/2u(t)+(1+et)v(2+cos2t)(1+v4/7)+67=0,cD7/3v(t)+(1+e3t)ln(1+u)(1+e2tsint)(2+u1/3)+117=0,u(0)=D3/2u(0)=D3/2u(1)=0,v(0)=D5/4v(0)=D5/4v(1)=0, (12)

    其中,0 < t < 1,n=2,2 < α=5/2, β=7/3 < 3,满足1+γα, 1+ δβ,而且

    f(t,v)=(1+et)v(2+cos2t)(1+v4/7)+67((t,v)[0,1]×[0,)),
    g(t,u)=(1+e3t)ln(1+u)(1+e2tsint)(2+u1/3)+117((t,v)[0,1]×[0,)).

    易知f, gC([0, 1]×[0, ∞), [0, ∞)),且

    |f(t,v)|0.683940v3/7+0.857143,
    |g(t,u)|21.085537u2/3+1.571429.

    定理2的条件均被满足,故耦合系统(12)在[0, 1]上至少有1个正解.

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出版历程
  • 收稿日期:  2019-07-08
  • 网络出版日期:  2021-03-21
  • 刊出日期:  2020-04-24

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